Consider the line which passes through the point P(-1, -5, 4), and which is parallel to the line x=1+5t y=2+5t z=3+5t. How do you find the point of intersection of this new line with each of the coordinate planes?

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Answer 1 · 2017-03-09T13:50:28

The reqd. pts. of intersection are,

#{(-5,-9,0)}, {(0,-4,5)} and {(4,0,9)}#.

We first find the eqn. of the new line, say #L.#

We note that the Direction Vector of the given line is,

# (5,5,5)=5(1,1,1)#

Since the line #L# is parallel to this given one, we may take, as its

direction vector, #vec l=(1,1,1).# Also, #P(-1,-5,4) in L.#

Hence, the Cartesian Parametric Eqns. of #L# are,

# {x-(-1)}/1={y-(-5)}/1=(z-4)/1=k, say, where, k in RR; or,#

# L : x=k-1, y=k-5, z=k+4, k in RR... ... ... ...(star)#

Now, to find the pt. of int. of #L# with #3# co-ordinate planes, namely,

#(1)# the XY co-ord. pln., #(2)# the YZ pln. and, #(3)# the XZ pln.,

let us recall that, their resp. eqns. are,

#(1) :z=0; (2) : x=0, & (3) : y=0.#

Therefore, to find, #L nn XY"-plane, we solve "(star) & (1).#

Sub.ing, #z=0# in #(star)#, we get,

#k=-4, so, x=k-1=-4-1=-5, &, y=-9.#

# rArr L nn XY-#plane=#{(-5,-9,0)}.#

Similarly, #L nn YZ"-plane="{(0,-4,5)}...[because, k=1], and, #

# L nn XZ"-plane="{(4,0,9)}...[because, k=5].#

Thus, the reqd. pts. of intersection are,

#{(-5,-9,0)}, {(0,-4,5)} and {(4,0,9)}#.

Enjoy Maths.!